# Resistance of a resistor and the energy and power dissipated

1. Sep 25, 2008

### isaaclimdc

Hi all,

I'm doing a lab now to determine the relationship between the resistance of a resistor (colour-coded ones) and the energy and power dissipated. What kind of relationship is expected between the variables R and P, or R and E?

The way I carried out the lab was to use different resistors (resistances known), and observe the change in the current using an ammeter in series. I observed that as resistance increases, current decreases. This is expected. Is power meant to be constantly varying, or is it meant to be constant?

Thanks for any help.

2. Sep 25, 2008

### jablonsky27

Power dissipated is proportional to resistance. So if you plot the two, should be a straight line.

3. Sep 25, 2008

### isaaclimdc

That is if P=I^2R right? But why is it that if P=V^2/R is used, you get a relationship that says otherwise?

4. Sep 25, 2008

### ZapperZ

Staff Emeritus
That makes no sense. Substitute V=IR into the first equation and you get the second. So they are identical! You cannot get a "... relationship that says otherwise.."

Zz.

5. Sep 25, 2008

### atyy

Yes, it's peculiar, in one case P increases with R, in the other case P decreases with R! I think it's something like this - say we connect a resistor R in series with a battery V.

Let's call:
P=I2R [Formula 1]
P=V2/R [Formula 2]

Ohm's law:
R=V/I, where R is a constant.

If we wish to test Formula 1 by increasing R while fixing I, we find that we cannot do this unless we also increase V - because Ohm's law says that if V is fixed and R increased, then I decreases. Since here we need to increase both R and V, Formulas 1 and 2 will agree, and both will predict that P increases with increasing R.

If we wish to test Formula 2 by increasing R while fixing V, we can do this quite easily since the battery puts out a fixed V. Ohm's law says that increasing R causes I to decrease, so both Formulas 1 and 2 will again agree, and both will predict that P decreases with increasing R.

6. Sep 25, 2008

### isaaclimdc

Yes, but why is it that the proportion between R and P are different in the two equations?

7. Sep 25, 2008

### ZapperZ

Staff Emeritus
Because they involved different variables as well!

Are we have problems with the physics, or the mathematics here?

Zz.

8. Sep 25, 2008

### isaaclimdc

Okay... but let's say I plot a R vs I^2 graph, and its an exponential curve with negative gradient. Does the integral of the function from my lowest R value to the highest i took equal the power?

It seems like power stays CONSTANT no matter what the resistance is! Since P=IV, and as R increases, I decreases, and the Terminal voltage hence increases (T.V. = emf - Ir).

Does this make sense?

9. Sep 25, 2008

### cabraham

The difference lies in the fact that with P=(I^2)*R, P increases with increasing R because the *current*, I, is stipulated constant. It is understood that raising R while holding I constant also raises V. Ultimately P=I*V.

With P=(V^2)/R, lowering R while holding V constant results in an increase in P because I also increases.

Generally, any one equation will not contain everything you should know. To fully grasp the way power is related to resistance one must consider both equations, V=I*R, and P=I*V. There are 4 varables, I, V, R, & P. If you know 2, you can compute the other 2.

Does this help?

Claude

10. Sep 25, 2008

### Staff: Mentor

It has been alluded to, but just to perhaps put it more succinctly: You can't always just fix one value and let the others vary. In this case, the first equation has two dependent variables. In other words, in a typical real world situation (such as a light with a rheostat on it), you cannot vary the resistance without also varying the amperage.

That is why I prefer not combining the equations until people understand how they work separately. Ie, learn how V=IR works in a real circuit, then apply that knowledge to the definition of power: P=VI.

11. Sep 25, 2008

### isaaclimdc

Yes, I agree with that, and that's why I have taken measurements for all three variables (R, I and V). However, I just need to know if P is expected to be constant or not? That is what is confusing me..

12. Sep 25, 2008

### atyy

If R is much larger than the terminal resistance r, then the voltage V across R is approximately constant (TV=emf). First plot your data and see if you can fit your formulas to the data. If you can, then you can be happy, I think. If you can't, then reconsider your assumptions and rederive the formulas for two resistors R and r in series.

13. Sep 25, 2008

### Staff: Mentor

If you want to know how power varies with a varying resistance and constant voltage, look at the equation that has those three terms in it and you tell us what kind of relationship it is.... I think you'll get it right. You're confusing yourself because you are looking at an equation that isn't describing what you are trying to test. The amperage is only measured as a way to find power. It is raw data, not the final data that you are trying to analyze.

14. Sep 27, 2008

### Redbelly98

Staff Emeritus
It depends on the nature of the circuit's power source, which you haven't told us about, so nobody can answer that question (yet).

Are you supplying a fixed voltage? Then P is inversely proportional to R. (P = V2/R)

Are you supply a fixed current? Then P is directly proportional to R. (P = I2R}

Are neither V nor I fixed? Then P has a more complicated dependence on R.